(i) CH₃ - OH : Does not exist as resonance hybrid due to absence of π-electrons.
(ii) R-CO NH₂ : Can exist as resonance hybrid due to the presence of non-bonding electrons on N and n-electrons on C = O bond.

(iii) CH₃-CH = CH-CH₂NH₂: Does not exist as resonance hybrid, because the lone pair on N-atom is not conjugated with n-electrons of the double bond

Optical isomerism:
Compounds having same physical and chemical property but differ only in the rotation of plane of the polarized light are known as optical isomers and the phenomenon is known as optical isomerism.
Some organic compounds such as glucose have the ability to rotate the plane of the plane polarized light and they are said to be optically active compounds and this property of a compound is called optical activity. The optical isomer, which rotates the plane of
the plane polarised light to the right or in cloclauisl direction is said to be dextrorotary (dexter means right) denoted by the sign (+), whereas the compound which rotates to the left or anticloclanrise is said to be leavo rotatory (leanues mean left) denoted by sign(-).
Dextrorotatory compounds are represented as 'd' or by sign (+) and lavorotatory compounds are ( - ) represented as 'l' or by sign (-).
Enantiomerism and optical activity: An optically active substance may exist in two or more isomeric forms which have same physical and chemical properties but differ in terms of direction of rotation of plane polarized light, such optical isomers which rotate the plane of polarized light with equal angle but in opposite direction are known as enantiomers and the phenomenon is knovrrn as enantiomerism. Isomers which are non-super impossible mirror images of each other are called enantiomers.
conditions for enantiomerism or optical isomerism:
A carbon atom whose tetra vaiency is satisfied by four different substituents (atoms or groups) is called a symmetric carbon or chiral carbon. It is indicated by an asterisk as C*. A molecule Possessing chiral carbon atom and non-super impossible to its own mirror image is said to be a chiral molecule or asymmetric, and the pioperty is called chirality or dissymmetry.

Weight of solute (W₂) = 20 mass percent of solution means 20 g of ethylene glycol
Weight of solvent (water) W₁ = 100 - 20 = 80 g
ΔT_f = K_f m
\(={K_f\times W_2\times 1000\over M_2\times W_1}\)
\(={1.86\times 20\times 1000\over 62\times 80}\)
= 7.5 K
The temperature at which the ice will begin to separate is the freezing of water after the addition of solute i.e 7.5 K lower than the normal freezing point of water (273 - 7.5K) = 265.5 K

If Q > K_c; the reaction will proceed in the reverse direction (ie) formation of reactants to proceed to reach equilibrium.

(i) Metallic hydrides are obtained by hydrogenation of metals and alloys in which hydrogen occupies the interstitial sites (voids). Hence, they are called interstitial hydrides.
(ii) The hydrides show properties similar to parent metals and hence they are also known as metallic hydrides.
(iii) They are mostly non-stoichiometric with variable composition (TiH_1.5-1.8 and PdH_0.6-0.8)
(iv) Some are relatively light, inexpensive and thermally unstable which makes them useful for hydrogen storage applications. Example, TiH₂, ZrH₂, ZnH₂.

This temperature below which a gas obey Joule Thomson effect is called inversion temperature (T_i).At the inversion temperature, no rise or fall in temperature of a gas occurs while expanding. But above the inversion temperature, the gas gets heated up when allowed to expand through a hole. It is related to Vander Waals equation by 
\({ T }_{ i }=\frac { 2a }{ Rb } \)

2(1) + x + 6(-1) = 0
2 + x - 6 = 0
x - 4 = 0
x = 4
Oxidation number of Fe in Na₂[Fe(CN)₆ ] is +4

(i) State function: A state function is a thermodynamic property of a system, which has a specific value for a given state and does not depend on the path (or manner) by which the particular state is reached.
Example: Pressure (P), Volume (V), Temperature(T)
(ii) Path functions: A path function is a thermodynamic property of the system whose value depends on the path by which the system changes from its initial to final states.
Example: Work (w), Heat (q).

(i) Hydrogen has the electronic configuration of 1s1 which resembles with ns¹ general valence shell configuration of alkali metals and shows similarity with them as follows:
1. It forms unipositive ion (H⁺) like alkali metals (Na⁺,K⁺,Cs⁺)
2. It forms halides (HX), oxides (H₂O), peroxides (H₂O₂) and sulphides (H₂S) like alkali metals (NaX, Na₂O, Na₂O₂, Na₂S)
3. It also acts as a reducing agent.
4. lt is an electro positive element
However, unlike alkali metals which have ionization energy ranging from 377 to 520 kJ mol^-1, the hydrogen has 1.314 kJ mol^-1 which is much higher than alkali metals.
Like the formation of halides (X ^-) from halogens, hydrogen also has a tendency to gain one electron to form hydride ion(H⁺) whose electronic configuration is similar to the noble gas, helium. However, the electron affinity of hydrogen is much less than that of halogen atoms. Hence, the tendency of hydrogen to form hydride ion is low compared to that of halogens to form the halide ions as evident from the following reactions:
\( 1 / 2 \mathrm{H}_{2}+\mathrm{e}^{-} \rightarrow \mathrm{H}^{-} \Delta \mathrm{H}=+36 \mathrm{kcal} \mathrm{mol}^{-1} \)
\(1 / 2 \mathrm{Br}_{2}+\mathrm{e}^{-} \rightarrow \mathrm{Br}^{-} \Delta \mathrm{H}=-55 \mathrm{kcal} \mathrm{mol}^{-1}\)
Since, hydrogen has similarities with alkali metals as well as the halogens; it is difficult to find the right position in the periodic table. However, in most of its compounds hydrogen exists in +1 oxidation state. Therefore, it is reasonable to place the hydrogen in group 1 along with alkali metals as shown in the latest periodic table published by IUPAC.

6Li + N₂ ⟶ 2Li₃ N

Erwin Schrodinger expressed the wave nature of electron in terms of a differential equation. This equation determines the change of wave function in space depending on the field of force in which the electron moves. The time independent Schrodinger equation can be expressed as,
\(\overset { \wedge }{ H } \psi =E\psi \)          .........(1)
Where \(\overset { \wedge }{ H } \) is called Hamiltonian operator, \(\psi \) is the wave function and is a function of position coordinates of the particle and is denoted as \(\psi \) (x, y, z) E is the energy of the system
\(\overset { \wedge }{ H } =\left[ \frac { { -h }^{ 2 } }{ 8{ \pi }^{ 2 } } \left( \frac { { \partial }^{ 2 }\psi }{ { \partial x }^{ 2 } } +\frac { { \partial }^{ 2 }\psi }{ { \partial y }^{ 2 } } +\frac { { \partial }^{ 2 }\psi }{ { \partial z }^{ 2 } } \right) +V \right] \)
can be written as
\(\left[ \frac { { -h }^{ 2 } }{ 8{ \pi }^{ 2 }m } \left( \frac { { \partial }^{ 2 }\psi }{ { \partial x }^{ 2 } } +\frac { { \partial }^{ 2 }\psi }{ { \partial y }^{ 2 } } +\frac { { \partial }^{ 2 }\psi }{ { \partial z }^{ 2 } } \right) +V\Psi \right] =E\Psi \)
Multiply by \(\frac { 8{ \pi }^{ 2 }m}{ { -h }^{ 2 } } \)and rearranging
\(\frac { { \partial }^{ 2 }\psi }{ { \partial x }^{ 2 } } +\frac { { \partial }^{ 2 }\psi }{ { \partial y }^{ 2 } } +\frac { { \partial }^{ 2 }\psi }{ { \partial z }^{ 2 } } +\frac { 8{ \pi }^{ 2 }m }{ { -h }^{ 2 } } (E-V)\Psi =0\)          ........(2)
The above Schrodinger wave equation does not contain time as a variable and is referred to as time independent Schrodinger wave equation. This equation can be solved only for certain values of E, the total energy. i.e. the energy of the system is quantised. The permitted total energy values are called eigen values and corresponding wave functions represent the atomic orbitals.

In the periodic table vertical columns are called groups.
Elements in the same vertical column possess similar number of electrons in the outer orbitals.
∴ Elements a and c belongs to group 18
Elements b and d belongs to group 13

(i) The decreasing order of bond angles is: NH₄⁺ > NH₃>NH₂^-
(ii) The N-atoms in all the three molecules are sp³ hybridised
(iii) The number of lone-pair of electrons present on nitrogen of NH₄⁺, NH₃ and NH₂^- are 0, 1 and 2.
(iv) Greater the number of lone pairs, greater is the repulsion and lesser is the bond angle.

ΔG⁰ = - 2.303 RT log k_c
= - 2.303 x 8.314 x 298 log (1.8 x 10^-7)
= - 38484 J mol^-1
= - 38.484 kJ mol-1

(i) Avogadro's law
Equal volumes of all the gases under the same conditions of temperature and pressure contain equal number of molecules.
V \(\infty \)  n (at constant P and T)
(ii) Gay-Lussac's law.
At constant volume, the pressure of a fixed amount of a gas is directly proportional to the temperature. P \(\infty \) T (constant V)

2Ca + O₂ ⟶ 2CaO

(i) The aromatic hydrocarbon Ⓐ is Benzene, C₆H₆.
(ii) Benzene reacts. with Cl₂ to give chlorobenzene (C₆H₅CI), as Ⓑ as the product.

(iii) Chlorobenzene Ⓑ acts with sodium hydroxide to give (C₆HsOH) phenol, © as the product.

(iv) Chlorobenzene reacts with ammonia to give Aniline, C₆H₅NH₂Ⓓ as the product.

Ionic bond is formed between K⁺ and CI ^- When the electronegativity difference between the two combining atoms is large, the least electronegative atom completely transfers one or more of its valence electrons to the other combining atom so that both atoms can attain the nearest inert gas electronic configuration. The complete transfer of electron leads to the formation of a cation and an anion. Both these ions are held together by the electrostatic attractive force which is known as ionic bond.
Let us consider the formation potassium chloride. The electronic configuration of potassium and chlorine are
Potassium (K) : [Ar] 4s¹
Chlorine (CI) : [Ne]3s², 3p⁵
Potassium has one electron in its valence shell and chlorine has seven electron in its valence shell. By loosing one electron potassium attains the inert gas electronic configuration of argon and becomes a unpositive cation (K⁺) and chlorine accepts this electron to become unnegative chloride ion (Cl^-) there by attaining the stable electronic conguration of argon. These two ions combine to form an ionic crystal in which they are held together by electrostatic attractive force. The energy required for the formation of one mole of K⁺ is 418.81 kJ (ionization energy) and the energy released during the formation of one mole of Cl- is -348.56 kJ (electron gain enthalpy). The sum of these two energies is positive. (70.25 kJ). However, during the formation of one mole potassium chloride crystal from its constituent ions, 718 kJ energy is released. This favours the formation of KCI and it stabilises.

(i) BOD and COD

(ii) Viable and non-viable particulate pollutants

Oxidation number method:
This method is based on the fact that
Number of electrons lost by atoms = Number of electrons gained by atoms
Steps to be followed while balancing Redox reactions by Oxidation Number method:
1. Write skeleton equation representing redox reaction
2. Write the oxidation number of atoms undergoing oxidation and reduction.
3. Calculate the increase or decrease in oxidation numbers per atom.
4, Make increase in oxidation number equal to decrease in oxidation number by multiplying the formula of oxidant and reductant by suitable numbers.
5. Balance the equation atomically on both sides except O and H atoms.
6. Balance oxygen atoms by adding required number of water molecules to the side deficient in oxygen atoms.
7. Add required number of H⁺ ions to the side deficient in hydrogen atom if the reaction is in acidic medium.
8. For reactions in basic medium, add H₂O molecules to the side deficient in hydrogen atoms and simultaneously add equal number of OH^- ions on the other side of the equation.
9. Finally, balance the equation by cancelling common species present on both sides of the equation.

Factors influencing ionization enthalpy:
(i) Size of the atom:
If the size of an atom is larger, the outermost electron shell from the nucleus is also larger and hence the outermost electrons experience lesser force of attraction. Hence it would be more easy to remove an electron from the outermost shell. Thus, ionization energy decreases with increasing atomic sizes.
Ionization enthalpy \(\infty{1\over Atomic \ size}\)
(ii) Magnitude of nuclear charge:
As the nuclear charge increases, the force of attraction between the nucleus and valence electrons also increases. So, more energy is required to remove a valence electron. Hence I.E increases with increase in nuclear charge.
Ionization enthalpy \(\alpha \ nuclear \ charge\)
(iii) Screening or shielding effect of the inner electrons:
The electrons of inner shells form a cloud of negative charge and this shields the outer electron from the nucleus. This screen reduces the coulombic attraction between the positive nucleus and the negative outer electrons. If screening effect increases, ionization energy decreases.
Ionization enthalpy  \(\infty{1\over Screening\ effects}\)
(iv) Penetrating power of subshells s, p, d, and f:
The s-orbital penetrate more closely to the nucleus as compared to p-orbitals. Thus, electrons in s-orbitals are more tightly held by the nucleus than electrons in p-orbitals. Due to this, more energy is required to remove a electron from an s-orbital as compared to a p-orbital. For the same value of 'n', the penetration power decreases in a given shell in the order.
s>p>d>f.
(v) Electronic configuration:
If the atoms of elements have either completely filled or exactly half filled electronic configuration, then the ionization energy increases.

(i) no. of electrons: 35 (given)
no. of protons : 35
(ii) Electronic configuration
1s² 2S² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁵
(iii) Last electron:

4P_x        4P_y     4p_z
last electron present in 4P_y orbital y
n = 4, l = 1 m₁ = either + 1 or -1 and s = -1/2

For 1 mole of an ideal gas,
PV=RT
P=4.1 atm
V=2lt.
PV=RT
T=\(\frac{PV}{R}=\frac{4.1atm\times2lit\times1mole}{0.082 lit atm K^{-1}mol^{-1}}\)
=100 K
\(\Delta S=\frac{q}{T}\)
\(\Delta S=\frac{3710J}{100K}=37.1 JK^{-1}\)
ΔS of expansion = 37.1 JK^-1.

half reaction \(\Rightarrow \) \({ S }_{ 2 }{ O }_{ 3 }^{ 2- }\longrightarrow { S }_{ 4 }{ O }_{ 6 }^{ 2- }\)
\({ I }_{ 2 }\longrightarrow { I }^{ - }\)
 

i) The fact that a small cation can stabilize a small anion and a large cation can stabilize a large anion explains the formation and stability of these oxides.
(ii) The Na⁺ ion is a larger cation and has a weak positive field around it and thus can stabilize a bigger peroxide ion, O₂^2- or [-O-O-]²- which is also surrounded by a weak negative field.
(iii) Similarly, the other ions K⁺, Rb⁺, Cs⁺ are still larger, having very weak positive field·
(iv) Thus these ions can stabilize a bigger superoxide O^2- anion and form super oxides·

∵ (V) < (ii) = (iv) < (iii) = (vi) < (i)
∴ 3p < 3d = 3d < 4p = 4p <4d